Introduction to Open Data Science - Course Project

About the project

Write a short description about the course and add a link to your GitHub repository here. This is an R Markdown (.Rmd) file so you should use R Markdown syntax.

# This is a so-called "R chunk" where you can write R code.
#version
date()

## [1] "Tue Nov 17 20:50:40 2020"

First time combining the use of the following tools for this Introduction to Open Data Science (IODS) course:

R, (version 4.0.3 (2020-10-10))
RStudio (version 1.3.1093)
Git (2.29.0)

Feeling very excited to embrace the future!

IODS project GitHub repository: https://github.com/mlammins/IODS-project

“By definition all scientists are data scientists. In my opinion, they are half hacker, half analyst, they use data to build products and find insights. It’s Columbus meet Columbo ― starry-eyed explorers and skeptical detectives.” ―Monica Rogati, Independent Data Science Advisor

2 Regression and model validation

Describe the work you have done this week and summarize your learning.

Describe your work and results clearly.
Assume the reader has an introductory course level understanding of writing and reading R code as well as statistical methods.
Assume the reader has no previous knowledge of your data or the more advanced methods you are using.

2.1 Introducing the dataset

The background of the data as described by the author:

Kimmo Vehkalahti: ASSIST 2014 - Phase 3 (end of Part 2), N=183 Course: Johdatus yhteiskuntatilastotieteeseen, syksy 2014 (Introduction to Social Statistics, fall 2014 - in Finnish), international survey of Approaches to Learning, made possible by Teachers’ Academy funding for KV in 2013-2015.

Data collected: 3.12.2014 - 10.1.2015/KV Data created: 14.1.2015/KV, in English 9.4.2015/KV,Florence,Italy Imputation 4.4.2015: only missing information in certain backgrounds, minimal amount of missing values imputed using Phases 1 and 2.

For more information, see https://www.mv.helsinki.fi/home/kvehkala/JYTmooc/JYTOPKYS3-meta.txt

learning2014 <- read.csv("./data/learning2014.csv") # reading the analysis data
dim(learning2014) # number of rows and columns

## [1] 166   7

str(learning2014) # type of data

## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : chr  "F" "M" "F" "M" ...
##  $ age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ attitude: num  3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ points  : int  25 12 24 10 22 21 21 31 24 26 ...

The learning dataset used in this exercise consists of 166 observations and 7 variables. Variables deep, stra and surf were calculated based on several Likert scale questions (from 1 to 5). Variable names and short descriptions:

gender - Gender: M (Male), F (Female)
age - Age (in years) derived from the date of birth
attitude - Global attitude toward statistics
deep - Tendency to deep learnign
stra - Tendency to strategic learning
surf - Tendency to surface learning
points - Exam points

2.2 Graphical overview and summary of the data

To get the idea of the data, let’s make a graphical summary of the variables with females (red) and males (blue):

library(ggplot2);
library(GGally);

## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2

p <- ggpairs(learning2014, mapping = aes(col=gender, alpha=0.3), lower = list(combo = wrap("facethist", bins = 20)))
p # graphical summary

summary(learning2014) # numerical summary

##     gender               age           attitude          deep      
##  Length:166         Min.   :17.00   Min.   :1.400   Min.   :1.583  
##  Class :character   1st Qu.:21.00   1st Qu.:2.600   1st Qu.:3.333  
##  Mode  :character   Median :22.00   Median :3.200   Median :3.667  
##                     Mean   :25.51   Mean   :3.143   Mean   :3.680  
##                     3rd Qu.:27.00   3rd Qu.:3.700   3rd Qu.:4.083  
##                     Max.   :55.00   Max.   :5.000   Max.   :4.917  
##       stra            surf           points     
##  Min.   :1.250   Min.   :1.583   Min.   : 7.00  
##  1st Qu.:2.625   1st Qu.:2.417   1st Qu.:19.00  
##  Median :3.188   Median :2.833   Median :23.00  
##  Mean   :3.121   Mean   :2.787   Mean   :22.72  
##  3rd Qu.:3.625   3rd Qu.:3.167   3rd Qu.:27.75  
##  Max.   :5.000   Max.   :4.333   Max.   :33.00

Taking a look at the graphical overview, the distribution for female (red) and male (blue) values seems to be relatively similar in all categories. Females seem to have a slightly higher values in surface learning (surf) and strategic learning (stra) and slighly lower values in attitude. The numerical summary of all data (not filtered by gender) shows that the majority of participants is young (under 30 years of age).

2.3 Fitting a regression model

Next, let’s test if there is correlation between attitude, strategic learning (stra) and surface learning tendency (surf) with the number of point obtained from the exam:

library(ggplot2)
# create a regression model with multiple explanatory variables
my_model <- lm(points ~ attitude + stra + surf, data = learning2014)
# print out a summary of the model
summary(my_model)

## 
## Call:
## lm(formula = points ~ attitude + stra + surf, data = learning2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.1550  -3.4346   0.5156   3.6401  10.8952 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.0171     3.6837   2.991  0.00322 ** 
## attitude      3.3952     0.5741   5.913 1.93e-08 ***
## stra          0.8531     0.5416   1.575  0.11716    
## surf         -0.5861     0.8014  -0.731  0.46563    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared:  0.2074, Adjusted R-squared:  0.1927 
## F-statistic: 14.13 on 3 and 162 DF,  p-value: 3.156e-08

Coefficients table gives an interpretation of the model. Intecept giving the “baseline” for the points in exam, it seems that attitude has the biggest correlation between exam points. If attitude rises one unit, then exam points increase by 3.4 units given that all other variables stay the same. The effect of stra and surf is below one. The importance of attitude can also be seen in the last column where the statistical significance of the coefficient is given. The three stars *** indicate high statistical significance, i.e. the coefficient differs from zero and thus has a relationship with the target variable. In general, the P-test value shown here tells the propability of the coefficient being zero.

Let’s drop surf (highest propability) to see if it improves the fitting:

# create a regression model with multiple explanatory variables
my_model2 <- lm(points ~ attitude + stra, data = learning2014)
# print out a summary of the model
summary(my_model2)

## 
## Call:
## lm(formula = points ~ attitude + stra, data = learning2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.6436  -3.3113   0.5575   3.7928  10.9295 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   8.9729     2.3959   3.745  0.00025 ***
## attitude      3.4658     0.5652   6.132 6.31e-09 ***
## stra          0.9137     0.5345   1.709  0.08927 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.289 on 163 degrees of freedom
## Multiple R-squared:  0.2048, Adjusted R-squared:  0.1951 
## F-statistic: 20.99 on 2 and 163 DF,  p-value: 7.734e-09

Leaving out surface learning increases the significance of the remaining variables (decreased values in P-value column). However, the value for strategic learning (stra) is still relatively high (debatable). As it is less than < 0.10, let’s keep it in the model. So we have found our final model.

2.4 Analyzing the fitted model

Our final model is thus

exam points = 8.97 + 3.47 x attitude + 0.91 x stra

This means that with one unit increase in attitude exam points increase by 3.47 points (stra being unchanged) and with one unit increase in strategic learning (stra) the exam points increase by 0.91. The baseline for exam points is 8.97 (the y-intercept). Now this is the systemic part (no error).

The numbers at the end of the multiple regression summary can be explained succinctly as:

Residual Standard Error: Standard deviation of residuals / errors of the regression model.
Multiple R-Squared: Percent of the variance of exam intact after subtracting the error of the model.
Adjusted R-Squared: Same as multiple R-Squared but takes into account the number of samples and variables.

So adjusted R-squared tells how well the model fits the data, i.e. the percentage of the dependent variable variation that the linear model explains (ranging between 0 and 1). The R-squared seen here (roughly 0.20) is quite low so there is probably some problematic patterns in the residual plots. At least the residuals are not very close to the regression line. That’s why you cannot rely on the R-squared number alone but a visual inspection is a must!

2.5 Model validity and diagnostic plots

# drawing diagnostic plots using the plot() function. Choose the plots 1, 2 and 5:
par(mfrow = c(2,2))
plot(my_model2, which=c(1,2,5))

A statistical model always include assumptions which describe the data generation process. In this linear regression case we assume

linearity, i.e. the target variable can be modelled as a linear combination of the model parameters
the errors are normally distributed, are not correlated and have constant variance (the size of the given error does not depend on the explanatory variables)

These assumptions can be checked through analyzing residuals.

Residuals vs. model predictions i.e. fitted values: for analyzing constant variance of errors. “Any pattern in the plot implies a problem with the assumptions.”
QQ-plot: for analyzing the normality of errors. “The better residuals are located on the identity line, the better the normality assumption holds.”
Leverage: for analyzing the impact of a single observation on the model. “The outliers have high impact.”

In residuals vs. fitted we can see the source of the low R-squared value: the residuals are not very close to the regression line (note y-axis scale). However, they seem to describe the trend in observations well since there is no noticeable pattern. QQ-plot also shows that the residuals are nicely located on the line, i.e. the normality assumption holds. Residuals vs. leverage shows some points at the right hand side of the plot but the x-axis scaling (max leverage value of 0.05) is still quite small, so no problems here either. All in all, it seems our multiple regression model nicely describes the data and all the assumptions hold.

With the model validated, now we could use our regression model to predict the target variable behavior!

3 Logistic regression

Describe the work you have done this week and summarize your learning.

Describe your work and results clearly.
Assume the reader has an introductory course level understanding of writing and reading R code as well as statistical methods.
Assume the reader has no previous knowledge of your data or the more advanced methods you are using.

3.1 Introducing the joined dataset

The dataset describes student achievement in secondary education of two Portuguese schools. The dataset is joined from two datasets regarding the performance in two distinct subjects: Mathematics (mat) and Portuguese language (por). The data attributes include student grades, demographic, social and school related features (and especially alcohol consumption) and it was collected by using school reports and questionnaires.

More information: https://archive.ics.uci.edu/ml/datasets/Student+Performance

alc <- read.csv("./data/alc.csv") # reading the analysis data
dim(alc) # number of rows and columns

## [1] 382  35

str(alc) # type of data

## 'data.frame':    382 obs. of  35 variables:
##  $ school    : chr  "GP" "GP" "GP" "GP" ...
##  $ sex       : chr  "F" "F" "F" "F" ...
##  $ age       : int  18 17 15 15 16 16 16 17 15 15 ...
##  $ address   : chr  "U" "U" "U" "U" ...
##  $ famsize   : chr  "GT3" "GT3" "LE3" "GT3" ...
##  $ Pstatus   : chr  "A" "T" "T" "T" ...
##  $ Medu      : int  4 1 1 4 3 4 2 4 3 3 ...
##  $ Fedu      : int  4 1 1 2 3 3 2 4 2 4 ...
##  $ Mjob      : chr  "at_home" "at_home" "at_home" "health" ...
##  $ Fjob      : chr  "teacher" "other" "other" "services" ...
##  $ reason    : chr  "course" "course" "other" "home" ...
##  $ nursery   : chr  "yes" "no" "yes" "yes" ...
##  $ internet  : chr  "no" "yes" "yes" "yes" ...
##  $ guardian  : chr  "mother" "father" "mother" "mother" ...
##  $ traveltime: int  2 1 1 1 1 1 1 2 1 1 ...
##  $ studytime : int  2 2 2 3 2 2 2 2 2 2 ...
##  $ failures  : int  0 0 2 0 0 0 0 0 0 0 ...
##  $ schoolsup : chr  "yes" "no" "yes" "no" ...
##  $ famsup    : chr  "no" "yes" "no" "yes" ...
##  $ paid      : chr  "no" "no" "yes" "yes" ...
##  $ activities: chr  "no" "no" "no" "yes" ...
##  $ higher    : chr  "yes" "yes" "yes" "yes" ...
##  $ romantic  : chr  "no" "no" "no" "yes" ...
##  $ famrel    : int  4 5 4 3 4 5 4 4 4 5 ...
##  $ freetime  : int  3 3 3 2 3 4 4 1 2 5 ...
##  $ goout     : int  4 3 2 2 2 2 4 4 2 1 ...
##  $ Dalc      : int  1 1 2 1 1 1 1 1 1 1 ...
##  $ Walc      : int  1 1 3 1 2 2 1 1 1 1 ...
##  $ health    : int  3 3 3 5 5 5 3 1 1 5 ...
##  $ absences  : int  5 3 8 1 2 8 0 4 0 0 ...
##  $ G1        : int  2 7 10 14 8 14 12 8 16 13 ...
##  $ G2        : int  8 8 10 14 12 14 12 9 17 14 ...
##  $ G3        : int  8 8 11 14 12 14 12 10 18 14 ...
##  $ alc_use   : num  1 1 2.5 1 1.5 1.5 1 1 1 1 ...
##  $ high_use  : logi  FALSE FALSE TRUE FALSE FALSE FALSE ...

The final joined dataset has 382 observations and 35 variables consisting of only unique individuals. The datasets were joined by using the 13 student identifier variables: “school”, “sex”, “age”, “address”, “famsize”, “Pstatus”, “Medu”, “Fedu”, “Mjob”, “Fjob”, “reason”, “nursery” and “internet”. Only students present in both datasets were kept. The variables not used for joining the two data have been combined by averaging (including the grade variables). A more detailed information about the variables is presented below (possible values in parenthesis):

1 school - student’s school (binary: ‘GP’ - Gabriel Pereira or ‘MS’ - Mousinho da Silveira)
2 sex - student’s sex (binary: ‘F’ - female or ‘M’ - male)
3 age - student’s age (numeric: from 15 to 22)
4 address - student’s home address type (binary: ‘U’ - urban or ‘R’ - rural)
5 famsize - family size (binary: ‘LE3’ - less or equal to 3 or ‘GT3’ - greater than 3)
6 Pstatus - parent’s cohabitation status (binary: ‘T’ - living together or ‘A’ - apart)
7 Medu - mother’s education (numeric: 0 - none, 1 - primary education (4th grade), 2 5th to 9th grade, 3 secondary education or 4 higher education)
8 Fedu - father’s education (numeric: 0 - none, 1 - primary education (4th grade), 2 5th to 9th grade, 3 secondary education or 4 higher education)
9 Mjob - mother’s job (nominal: ‘teacher’, ‘health’ care related, civil ‘services’ (e.g. administrative or police), ‘at_home’ or ‘other’)
10 Fjob - father’s job (nominal: ‘teacher’, ‘health’ care related, civil ‘services’ (e.g. administrative or police), ‘at_home’ or ‘other’)
11 reason - reason to choose this school (nominal: close to ‘home’, school ‘reputation’, ‘course’ preference or ‘other’)
12 nursery - attended nursery school (binary: yes or no)
13 internet - Internet access at home (binary: yes or no)
14 guardian - student’s guardian (nominal: ‘mother’, ‘father’ or ‘other’)
15 traveltime - home to school travel time (numeric: 1 - <15 min., 2 - 15 to 30 min., 3 - 30 min. to 1 hour, or 4 - >1 hour)
16 studytime - weekly study time (numeric: 1 - <2 hours, 2 - 2 to 5 hours, 3 - 5 to 10 hours, or 4 - >10 hours)
17 failures - number of past class failures (numeric: n if 1<=n<3, else 4)
18 schoolsup - extra educational support (binary: yes or no)
19 famsup - family educational support (binary: yes or no)
20 paid - extra paid classes within the course subject (Math or Portuguese) (binary: yes or no)
21 activities - extra-curricular activities (binary: yes or no)
22 higher - wants to take higher education (binary: yes or no)
23 romantic - with a romantic relationship (binary: yes or no)
24 famrel - quality of family relationships (numeric: from 1 - very bad to 5 - excellent)
25 freetime - free time after school (numeric: from 1 - very low to 5 - very high)
26 goout - going out with friends (numeric: from 1 - very low to 5 - very high)
27 Dalc - workday alcohol consumption (numeric: from 1 - very low to 5 - very high)
28 Walc - weekend alcohol consumption (numeric: from 1 - very low to 5 - very high)
29 health - current health status (numeric: from 1 - very bad to 5 - very good)
30 absences - number of school absences (numeric: from 0 to 93)

The grades G1, G2 and G3 are related to the course subject, Math or Portuguese. Variables alc_use and high_use were added to the original datasets:

31 G1 - first period grade (numeric: from 0 to 20)
32 G2 - second period grade (numeric: from 0 to 20)
33 G3 - final grade (numeric: from 0 to 20, output target)
34 alc_use - average of Dalc and Walc (numeric: from 1 - very low to 5 - very high)
35 high_use - high alcohol use status (binary: TRUE (if alc_use is higher than 2) or FALSE)

3.2 Purpose of the analysis and study hypothesis

The purpose of the analysis is to study the relationships between high/low alcohol consumption and some of the other variables in the data. Out of the many variables present, parent’s cohabitation status, mother’s education, quality of family relationships and the number of school absences were chosen. Thus the study hypothesis are as follows:

Hypothesis 1: If the parents are living apart (Pstatus=‘A’), the alcohol consumption (high_use) is higher
Hypothesis 2: The higher the mother’s education (Medu), the lower the alcohol consumption (high_use)
Hypothesis 3: The higher the quality of family relationships (famrel), the lower the alcohol consumption (high_use)
Hypothesis 4: The higher the number of school absences (absences), the higher the alcohol consumption (high_use)

Note! Now the target variable high_use is a binary variable (TRUE=1, FALSE=0) so we must use logistic regression.

3.3 Numerical and graphical exploration

Before making logistic regression, let’s explore the distribution of chosen variables and their connection to the target variable numerically and graphically. First, let’s take a look at variable distributions:

library(dplyr);
# use dplyr to make a smaller dataset and include all chosen variables
alc_test <- select(alc, Pstatus, Medu, famrel, absences, alc_use, high_use)
# change Pstatus from char to factor
alc_test$Pstatus <- as.factor(alc_test$Pstatus)
# numerical summary of chosen variables
summary(alc_test[-c(5,6)])

##  Pstatus      Medu           famrel         absences   
##  A: 38   Min.   :0.000   Min.   :1.000   Min.   : 0.0  
##  T:344   1st Qu.:2.000   1st Qu.:4.000   1st Qu.: 1.0  
##          Median :3.000   Median :4.000   Median : 3.0  
##          Mean   :2.806   Mean   :3.937   Mean   : 4.5  
##          3rd Qu.:4.000   3rd Qu.:5.000   3rd Qu.: 6.0  
##          Max.   :4.000   Max.   :5.000   Max.   :45.0

# graphical exploration of variable distribution
par(mfrow = c(2,2))
#
barplot(table(alc_test$Pstatus), main="Distribution of Pstatus")
barplot(table(alc_test$Medu), main="Distribution of Medu")
barplot(table(alc_test$famrel), main="Distribution of famrel")
barplot(table(alc_test$absences), main="Distribution of absences")

From the data we can see that the vast majority of participants are living together with their parents (T). Also, most of the participants have good family relations (famrel=4) and the number of absences is relatively small (75 % less than 6). Mother’s education is almost evenly distributed among variable values not counting zero.

Out of curiosity, let’s see how the alcohol consumption is distributed.

summary(alc_test[c(5,6)])

##     alc_use       high_use      
##  Min.   :1.000   Mode :logical  
##  1st Qu.:1.000   FALSE:268      
##  Median :1.500   TRUE :114      
##  Mean   :1.889                  
##  3rd Qu.:2.500                  
##  Max.   :5.000

par(mfrow = c(1,2))
barplot(table(alc_test$alc_use), main="Distribution of alc_use")
barplot(table(alc_test$high_use), main="Distribution of high_use")

It seems that only about one third of the participants use alcohol in high volumes. To better grasp the situation, here also the alcohol use (alc_use, numeric from 1 to 5) is shown instead of only the binary variable high use (high_use, TRUE if alc_use > 2).

Now let’s see what is the relation of our chosen variables to alcohol consumption. Here also the numerical alc_use is used.

par(mfrow = c(2,2))
boxplot(alc_use ~ Pstatus, data = alc)
boxplot(alc_use ~ Medu, data=alc)
boxplot(alc_use ~ famrel, data=alc)
boxplot(alc_use ~ absences, data=alc)

These boxplots give a rough idea of the relationships between variables and the target variable:

Pstatus vs. alc_use: Roughly the same Q1 to Q3 area, but the median is smaller in A group. This is contrary to what was hypothesized.
Medu vs. alc_use: Certainly group 0 has the greatest median for alc_use and the trend seems to go down as Medu increases, but group 3 and 4 do have big variability in values (Q1-Q3 cover a large area).
famrel vs. alc_use: Nothing conclusive on the hypothesis. Excellent family relations (famrel=5) seems to protect from high alcohol consumption (smaller median, smaller Q1-Q3 area).
absences vs. alc_use: Alcohol use seems to increase with absences which supports the hypothesis.

3.4 Logistic regression - the model

Time to form a mathematically rigorous model using logistic regression! Note that Pstatus will be addressed here as factor. To summarize:

Target variable = high_use

Chosen variables: Pstatus, Medu, famrel, absences

m <- glm(high_use ~ Pstatus+Medu+famrel+absences, data = alc_test, family ="binomial")
# print out a summary of the model
summary(m)

## 
## Call:
## glm(formula = high_use ~ Pstatus + Medu + famrel + absences, 
##     family = "binomial", data = alc_test)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2741  -0.8107  -0.7076   1.1985   1.8620  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -0.453190   0.697544  -0.650 0.515890    
## PstatusT     0.168029   0.397078   0.423 0.672176    
## Medu        -0.008959   0.107430  -0.083 0.933542    
## famrel      -0.243649   0.124088  -1.964 0.049585 *  
## absences     0.088049   0.022951   3.836 0.000125 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 443.46  on 377  degrees of freedom
## AIC: 453.46
## 
## Number of Fisher Scoring iterations: 4

The most important insight from the model summary is the coefficient section. Using the estimated coefficients, the obtained model is

high_use $= -0.45+0.17 \times PstatusT-0.01 \times Medu-0.24 \times famrel+0.09 \times absences$

Since the target valuable high_use is binary, it has only values 0 (as FALSE or “failure”) and 1 (as TRUE or "success) in modelling sense. The coefficients of the fitted model can be thus interpreted as

PstatusP: if Pstatus=A, then PstatusT=0 and the coefficient is -0.45 meaning parents living apart actually lessens the probability of high alcohol use. If Pstatus=T, then the effect is -0.45+0.17=-0.28 which is still negative but not as much as with Pstatus=A.
Medu: higher level of mother’s education lessens the probability for high alcohol use but the effect is quite negligible.
famrel: negative coefficient means high value for famrel lessens the probability of high_use.
absences: positive value for coefficient means that increased number of absences increases the probability of high_use

Of the variables, only famrel and absences were found to be statistically significant on 0.05 and 0.001 levels, respectively.

The model coefficients can also be interpreted as odds ratio.

Odds: the ratio of expected “successes” to “failures”, i.e. $\frac{p}{1-p}$ with value ranging from 0 to infinity.

So higher odds correspond to a higher probability of success. They are an alternative way of expressing probabilities. Let’s calculate the odds ratios and confidence intervals for the coefficients.

# print out the coefficients of the model
coef(m)

##  (Intercept)     PstatusT         Medu       famrel     absences 
## -0.453190280  0.168028965 -0.008958589 -0.243649169  0.088049068

# compute odds ratios (OR)
OR <- coef(m) %>% exp
# compute confidence intervals (CI)
CI <- confint(m) %>% exp
# print out the odds ratios with their confidence intervals
cbind(OR, CI)

##                    OR     2.5 %   97.5 %
## (Intercept) 0.6355972 0.1580208 2.462028
## PstatusT    1.1829709 0.5566928 2.673024
## Medu        0.9910814 0.8032307 1.224977
## famrel      0.7837626 0.6137127 1.000047
## absences    1.0920417 1.0461954 1.144793

Odds ratio can be used to quantify the relationship between our variable and the target variable. Odds higher than 1 mean that the variable is positively associated with “success”. The odds ratio of our variables can be interpreted as

Pstatus: is connected to intercept and PstatusT as described above. Now the odds ratio for intercept is less than 1 meaning negative association with high alcohol use. On the other hand, PstatusT has positive association (greater than 1). But both have a large 95 % CI in which number 1 is included, so the significance is not strong.
Medu: the variable is very close to 1 (neither positive nor negative association) with 95 % CI including the number 1 so no significance.
famrel: less than 1 with 95 % CI upper end just barely greater than 1. Thus has a somewhat significant negative association with the target variable.
absences: greater than 1 and 95 % CI interval entirely above 1. Positive association with strong significance.

Comparison of the results with our hypotheses:

Hypothesis 1: Pstatus=A implies high_use. The results are completely contrary; parents living together has a slightly bigger probability of high alcohol use than parents living apart.
Hypothesis 2: The higher Medu implies no high_use. The results show that while high mother’s education does correlate with lower alcohol use, the effect is negligible.
Hypothesis 3: Higher famrel implies no high_use. This hypothesis holds; higher quality of family relations does imply lower alcohol consumption.
Hypothesis 4: Higher absences implies high_use. The hypothesis hold. This is, in fact, the most significant independent variable in the model.

3.5 Predictive power of the model

Let’s improve our model by discarding the most insignificant variables, Pstatus and Medu. Now the final model has only variables famrel and absences.

# fix the model
m2 <- glm(high_use ~ famrel+absences, data = alc_test, family ="binomial")
# print out a summary of the model
summary(m2)

## 
## Call:
## glm(formula = high_use ~ famrel + absences, family = "binomial", 
##     data = alc_test)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.3139  -0.8028  -0.7125   1.2100   1.8605  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -0.33027    0.50783  -0.650 0.515461    
## famrel      -0.24109    0.12365  -1.950 0.051211 .  
## absences     0.08668    0.02270   3.819 0.000134 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 443.66  on 379  degrees of freedom
## AIC: 449.66
## 
## Number of Fisher Scoring iterations: 4

It seems that the significance of variables famrel and absences has increased slightly compared to the original model. However, the Pr(>|z|) values are almost the same and we have a simpler model than before, so this is an improvement.

Let’s use the improved model and predict high_use values of an individual.

# predict() the probability of high_use
probabilities <- predict(m2, type = "response")
# add the predicted probabilities to 'alc_test'
alc_test <- mutate(alc_test, probability = probabilities)
# use the probabilities to make a prediction of high_use
alc_test <- mutate(alc_test, prediction = probabilities>0.5)
# see the last ten original classes, predicted probabilities, and class predictions
select(alc_test, famrel,absences, high_use, probability, prediction) %>% tail(10)

##     famrel absences high_use probability prediction
## 373      4        0    FALSE   0.2150733      FALSE
## 374      5        7     TRUE   0.2831342      FALSE
## 375      5        1    FALSE   0.1901522      FALSE
## 376      4        6    FALSE   0.3154941      FALSE
## 377      5        2    FALSE   0.2038593      FALSE
## 378      4        2    FALSE   0.2457776      FALSE
## 379      2        2    FALSE   0.3454525      FALSE
## 380      1        3    FALSE   0.4227907      FALSE
## 381      2        4     TRUE   0.3856256      FALSE
## 382      4        2     TRUE   0.2457776      FALSE

# tabulate the target variable versus the predictions
select(alc_test, high_use, prediction) %>% table()

##         prediction
## high_use FALSE TRUE
##    FALSE   259    9
##    TRUE    101   13

library(ggplot2)
# initialize a plot of 'high_use' versus 'probability' in 'alc_test'
g <- ggplot(alc_test, aes(x =probability, y = high_use, col=prediction))

# define the geom as points and draw the plot
g+geom_point()

# tabulate the target variable versus the predictions
table(high_use = alc_test$high_use, prediction = alc_test$prediction) %>% prop.table() %>% addmargins()

##         prediction
## high_use      FALSE       TRUE        Sum
##    FALSE 0.67801047 0.02356021 0.70157068
##    TRUE  0.26439791 0.03403141 0.29842932
##    Sum   0.94240838 0.05759162 1.00000000

# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc_test$high_use, prob = alc_test$probability)

## [1] 0.2879581

To interpret the result, the model ends up predicting wrong 29 % of the time. There is much room for improvement here although it is better than just guessing (50-50 chance). So the model is definitely better than nothing.

3.6 10-fold cross-validation of the model

# K-fold cross-validation
library(boot)
cv <- cv.glm(data = alc_test, cost = loss_func, glmfit = m2, K = 10)

# average number of wrong predictions in the cross validation
cv$delta[1]

## [1] 0.2931937

The prediction error obtained here is larger than in the model introduced in DataCamp (prediction error of 0.26). Choosing more significant variables in the model would improve the predictions.

4 Clustering and classification

Describe the work you have done this week and summarize your learning.

Describe your work and results clearly.
Assume the reader has an introductory course level understanding of writing and reading R code as well as statistical methods.
Assume the reader has no previous knowledge of your data or the more advanced methods you are using.

4.1 Introducing the dataset

This chapter’s dataset consists of housing values in suburbs of Boston (the Boston data from the MASS package).

# access the MASS package
library(MASS)

## 
## Attaching package: 'MASS'

## The following object is masked from 'package:dplyr':
## 
##     select

# load the data
data("Boston")
# explore the dataset
str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

The dataset contains 506 observations in 14 variables:

crim – per capita crime rate by town (numeric: from 0.00632 to 88.9762)
zn – proportion of residential land zoned for lots over 25,000 sq.ft. (numeric: from 0 to 100)
indus – proportion of non-retail business acres per town (numeric: from 0.46 to 27.74)
chas – Charles River dummy variable (binary: 1 if tract bounds river, 0 otherwise)
nox – nitrogen oxides concentration (parts per 10 million) (numeric: from 0.385 to 0.871)
rm – average number of rooms per dwelling (numeric: from 3.561 to 8.78)
age – proportion of owner-occupied units built prior to 1940 (numeric: 2.9 to 100)
dis – weighted mean of distances to five Boston employment centres (numeric: from 1.13 to 12.127)
rad – index of accessibility to radial highways (numeric: from 1 to 24)
tax – full-value property-tax rate per $10,000 (numeric: from 187 to 711)
ptratio – pupil-teacher ratio by town (numeric: from 12.6 to 22)
black – 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town (numeric: from 0.32 to 396.9)
lstat – lower status of the population (percent) (numeric: from 1.73 to 37.97)
medv – median value of owner-occupied homes in $1000s (numeric: from 5 to 50)

4.2 Graphical overview and summary

Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them. (0-2 points)

library(dplyr)
library(corrplot)

## corrplot 0.84 loaded

# explore the dataset
str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

pairs(Boston)

# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) 

# print the correlation matrix
cor_matrix %>% round(digits=2)

##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00

# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex=0.6)

4.3 Standardizing

Standardize the dataset and print out summaries of the scaled data. How did the variables change? Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). Use the quantiles as the break points in the categorical variable. Drop the old crime rate variable from the dataset. Divide the dataset to train and test sets, so that 80% of the data belongs to the train set. (0-2 points)

# center and standardize variables
boston_scaled <- scale(Boston)

# summaries of the scaled variables
summary(boston_scaled)

##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865

# class of the boston_scaled object
class(boston_scaled)

## [1] "matrix" "array"

# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)

# MASS, Boston and boston_scaled are available

# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins

##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610

# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE)

# look at the table of the new factor crime


# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

# boston_scaled is available

# number of rows in the Boston dataset 
n <- nrow(Boston)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

4.4 Fitting LDA

Fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Draw the LDA (bi)plot. (0-3 points)

# MASS and train are available

# linear discriminant analysis
lda.fit <- lda(crime ~., data = train)

# print the lda.fit object
lda.fit

## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
## [-0.419,-0.411]  (-0.411,-0.39] (-0.39,0.00739]  (0.00739,9.92] 
##       0.2524752       0.2524752       0.2475248       0.2475248 
## 
## Group means:
##                          zn      indus        chas        nox         rm
## [-0.419,-0.411]  0.88209581 -0.8508129 -0.07933396 -0.8473266  0.4062723
## (-0.411,-0.39]  -0.08789431 -0.2471167  0.03646311 -0.5668584 -0.1219210
## (-0.39,0.00739] -0.39119540  0.2475560  0.23949396  0.4270446  0.1412371
## (0.00739,9.92]  -0.48724019  1.0171519 -0.07547406  1.0810101 -0.4294293
##                        age        dis        rad        tax     ptratio
## [-0.419,-0.411] -0.8895773  0.8539187 -0.6767393 -0.7446541 -0.39472273
## (-0.411,-0.39]  -0.3148318  0.3406919 -0.5393736 -0.4415848 -0.06323769
## (-0.39,0.00739]  0.4275874 -0.3913932 -0.4087862 -0.2814048 -0.24967538
## (0.00739,9.92]   0.8214577 -0.8564107  1.6377820  1.5138081  0.78037363
##                       black       lstat         medv
## [-0.419,-0.411]  0.38540909 -0.75681126  0.498189377
## (-0.411,-0.39]   0.31765298 -0.14996459 -0.006125378
## (-0.39,0.00739]  0.02127577  0.01091389  0.179099166
## (0.00739,9.92]  -0.78990130  0.88686209 -0.694001749
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## zn       0.09897731  0.59170971 -0.94388854
## indus    0.01873827 -0.18041635  0.24891005
## chas    -0.08449136 -0.04876776  0.14105589
## nox      0.38623886 -0.86950189 -1.37536470
## rm      -0.10911244 -0.17483774 -0.18615528
## age      0.21962872 -0.32251356  0.07139725
## dis     -0.07872999 -0.27847459  0.19411354
## rad      3.19380274  1.02600365 -0.23873768
## tax     -0.02427280 -0.04481996  0.76711044
## ptratio  0.14009292 -0.08423752 -0.35942380
## black   -0.12605515  0.06675495  0.14801674
## lstat    0.28139797 -0.29464529  0.30966491
## medv     0.23977775 -0.45654221 -0.22862357
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9494 0.0374 0.0132

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results
plot(lda.fit, dimen = 2, col=classes, pch=classes)
lda.arrows(lda.fit, myscale = 1)

4.5 Predict

Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then predict the classes with the LDA model on the test data. Cross tabulate the results with the crime categories from the test set. Comment on the results. (0-3 points)

# lda.fit, correct_classes and test are available

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)

##                  predicted
## correct           [-0.419,-0.411] (-0.411,-0.39] (-0.39,0.00739] (0.00739,9.92]
##   [-0.419,-0.411]              21              4               0              0
##   (-0.411,-0.39]                6             14               4              0
##   (-0.39,0.00739]               1             12              12              1
##   (0.00739,9.92]                0              0               0             27

4.6 Distances and clustering

Reload the Boston dataset and standardize the dataset (we did not do this in the Datacamp exercises, but you should scale the variables to get comparable distances). Calculate the distances between the observations. Run k-means algorithm on the dataset. Investigate what is the optimal number of clusters and run the algorithm again. Visualize the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results. (0-4 points)

# load MASS and Boston
library(MASS)
data('Boston')

# euclidean distance matrix
dist_eu <- dist(Boston)

# look at the summary of the distances
summary(dist_eu)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.119  85.624 170.539 226.315 371.950 626.047

# manhattan distance matrix
dist_man <- dist(Boston,method="manhattan")

# look at the summary of the distances
summary(dist_man)

##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
##    2.016  149.145  279.505  342.899  509.707 1198.265

4.7 k-means

Bonus: Perform k-means on the original Boston data with some reasonable number of clusters (> 2). Remember to standardize the dataset. Then perform LDA using the clusters as target classes. Include all the variables in the Boston data in the LDA model. Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution). Interpret the results. Which variables are the most influencial linear separators for the clusters? (0-2 points to compensate any loss of points from the above exercises)

# Boston dataset is available
library(ggplot2)

# k-means clustering
km <-kmeans(Boston, centers = 3)

# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)

# MASS, ggplot2 and Boston dataset are available
set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

# k-means clustering
km <-kmeans(Boston, centers = 2)

# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)

4.8 Bonus

Super-Bonus: Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.

model_predictors <- dplyr::select(train, -crime)

# check the dimensions
dim(model_predictors)

## [1] 404  13

dim(lda.fit$scaling)

## [1] 13  3

# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

Next, install and access the plotly package. Create a 3D plot (Cool!) of the columns of the matrix product by typing the code below.

library(plotly)

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:MASS':
## 
##     select

## The following object is masked from 'package:ggplot2':
## 
##     last_plot

## The following object is masked from 'package:stats':
## 
##     filter

## The following object is masked from 'package:graphics':
## 
##     layout

# Note! To install plotly in Linux, remember to install libcurl from terminal.
# * deb: libcurl4-openssl-dev (Debian, Ubuntu, etc)
# * rpm: libcurl-devel (Fedora, CentOS, RHEL)
# * csw: libcurl_dev (Solaris)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')

## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.

Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set. Draw another 3D plot where the color is defined by the clusters of the k-means. How do the plots differ? Are there any similarities? (0-3 points to compensate any loss of points from the above exercises)

(more chapters to be added similarly as we proceed with the course!)

IODS course project

Marko Lamminsalo

Introduction to Open Data Science - Course Project

About the project

2 Regression and model validation

2.1 Introducing the dataset

2.2 Graphical overview and summary of the data

2.3 Fitting a regression model

2.4 Analyzing the fitted model

2.5 Model validity and diagnostic plots

3 Logistic regression

3.1 Introducing the joined dataset

3.2 Purpose of the analysis and study hypothesis

3.3 Numerical and graphical exploration

3.4 Logistic regression - the model

3.5 Predictive power of the model

3.6 10-fold cross-validation of the model

4 Clustering and classification

4.1 Introducing the dataset

4.2 Graphical overview and summary

4.3 Standardizing

4.4 Fitting LDA

4.5 Predict

4.6 Distances and clustering

4.7 k-means

4.8 Bonus